A constant force of magnitude 82.0 N is applied to a 3.45 kg shoe box at angle p

Question

A constant force of magnitude 82.0 N is applied to a 3.45 kg shoe box at angle phi = 61.0°, causing the box to move up a frictionless ramp at constant speed. How much work is done on the box by F when the box has moved through vertical distance h = 0.130 m?

So I figured that the vertical component of the work exerted by the force must be F*h*cos(phi) since the force is not only at an angle, but an incline as well. Since gravity is a factor, I subtracted m*g*h*sin(phi), the work on gravity in this case due to the position of the object, from the above value. I tried this on a practice version and while it was close to the right answer, I couldn't get the right answer.

I'm not quite sure what I'm doing wrong here, but I feel like I have the right idea to solve this.

Explanation / Answer

W= Force x height x cos (phi)
I tried to draw a FBD but it doesnt load....so....

image an incline at 61 degrees with an applied force to a box up the ramp of 82N

and perpendicular from the box to the ground is another applied force due to gravity of

3.45kg * 9.8m/s^2 = 33.81 N

When the force acts perpendicular to the displacement then there is no work being done

thus only the component of the force in the same direction as the displacement (Fcos) does work

and this is canceled out by the normal force!

anyway to get to the point displacement is the hypotenuse

A=hyp. C=vert. distance C=A sin(phi)... C/sin (phi) = A

try working with A's distance (hyp.) instead on one of your practice problems!!

I really hope this helps...source: was special case right triangles..please rate

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